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Triangle read by rows, obtained from triangle A011973 by reading that array from right to left along the irregular paths shown in the figure.
4

%I #35 Jul 07 2024 00:54:24

%S 1,1,1,2,1,1,3,1,1,3,4,1,4,6,5,1,1,10,10,6,1,1,5,20,15,7,1,6,15,35,21,

%T 8,1,1,21,35,56,28,9,1,1,7,56,70,84,36,10,1,8,28,126,126,120,45,11,1,

%U 1,36,84,252,210,165,55,12,1,1,9,120,210,462,330,220,66,13,1

%N Triangle read by rows, obtained from triangle A011973 by reading that array from right to left along the irregular paths shown in the figure.

%C The successive rows have lengths 1,2,2; 3,4,4; 5,6,6; 7,8,8; ...

%C Sum of row n is A005314(n).

%C Old definition was: "Triangle of falling diagonals of A011973 (with rows displayed as centered text)."

%H N. J. A. Sloane, <a href="/A224838/a224838.pdf">Construction of present triangle by reading triangle A011973 from right to left along the paths indicated.</a>

%F r(n) = binomial(n-floor((4n+15-6k+(-1)^k)/12), n-floor((4n+15-6k+(-1)^k)/12)-floor((2n-1)/3)+k-1), k = 1..floor((2n+2)/3).

%F R(n) = binomial(n-floor((k+1)/2), n-floor((3k-1)/2)), k = 1..floor((2n+2)/3), gives the terms of each row in reverse order.

%e First 11 rows of the triangle:

%e 1;

%e 1, 1;

%e 2, 1;

%e 1, 3, 1;

%e 1, 3, 4, 1;

%e 4, 6, 5, 1;

%e 1, 10, 10, 6, 1;

%e 1, 5, 20, 15, 7, 1;

%e 6, 15, 35, 21, 8, 1;

%e 1, 21, 35, 56, 28, 9, 1;

%e 1, 7, 56, 70, 84, 36, 10, 1;

%t Table[Reverse[Table[Binomial[n - Floor[(k + 1)/2], n - Floor[(3 k - 1)/2]], {k, Floor[(2 n + 2)/3]}]], {n, 13}] (* _T. D. Noe_, Jul 25 2013 *)

%t Column[Table[Binomial[n - Floor[(4 n + 15 - 6 k + (-1)^k)/12], n - Floor[(4 n + 15 - 6 k + (-1)^k)/12] - Floor[(2 n - 1)/3] + k - 1], {n, 1, 25}, {k, 1, Floor[(2 n + 2)/3]}]] (* _John Molokach_, Jul 25 2013 *)

%Y Cf. A005314, A227300, A001973, A000045, A004396.

%K nonn,tabf

%O 1,4

%A _John Molokach_, Jul 21 2013

%E Entry revised by _N. J. A. Sloane_, Jul 07 2024